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MATHEMATICS COMPETITIONS

journal of the

WORLD FEDERATION OF NATIONAL MATHEMATICS COMPETITIONS

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AMT Publ ish ing

volume 26 number 1 2013

MATHEMATICS COMPETITIONS JournAl of The World federATion of nATionAl MATheMATics coMpeTiTions

(ISSN 1031 – 7503)Published biannually byAMT publ ish ing AusTrAl iAn MATheMAT ics TrusT

univers iTy of cAnberrA locked bAg 1cAnberrA gpo AcT 2601AusTrAl iA

With significant support from the UK Mathematics Trust.

Articles (in English) are welcome.

Please send articles to:

The EditorMathematics CompetitionsWorld Federation of National Mathematics CompetitionsUniversity of Canberra Locked Bag 1Canberra GPO ACT 2601Australia

Fax:+61-2-6201-5052

or

Dr Jaroslav ŠvrcekDept. of Algebra and GeometryFaculty of SciencePalacký University Tr. 17. Listopadu 12Olomouc772 02Czech Republic

Email: [email protected]

©2012 AMT Publishing, AMTT Limited ACN 083 950 341

MATheMATics coMpeTiTions voluMe 26 nuMber 1 2013

CONTENTS PAGE

WFNMC Committee 1

From the President 4

From the Editor 6

What “Problem Solving” Ought to Mean and how Combinatorial Geometry Answers this Question: Divertismento in Nine Movements 8

Alexander Soifer (United States of America)

Mathematics Competitions and Gender Issues: a Case of the Virtual Marathon 23

Mark Applebaum (Israel), Margo Kondratieva (Canada) and Viktor Freiman (Canada)

The Good, the Bad and the Pleasure (not Pressure!) of Mathematics Competitions 41

Siu Man Keung (Hong Kong)

The Twin Towers of Hanoi 59 Jack Chen, Richard Mah and Steven Xia (Canada)

Tournament of Towns 69 Andy Liu (Canada)

ˇ

Mathematics Competitions Vol 26 No 1 2013

World Federation of National MathematicsCompetitions

Executive

President: Professor Alexander SoiferUniversity of ColoradoCollege of Visual Arts and SciencesP.O. Box 7150 Colorado SpringsCO 80933-7150USA

Senior Vice President: Professor Kiril BankovSofia University St. Kliment OhridskiSofiaBULGARIA

Vice Presidents: Dr. Robert GeretschlagerBRG KeplerKeplerstrasse 18020 GrazAUSTRIA

Professor Ali RejaliIsfahan University of Technology8415683111 IsfahanIRAN

Publications Officer: Dr Jaroslav SvrcekDept. of Algebra and GeometryPalacky University, OlomoucCZECH REPUBLIC

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Secretary: Sergey DorichenkoSchool 179MoscowRUSSIA

ImmediatePast President &Chairman,Awards Committee:

Professor Marıa Falk de LosadaUniversidad Antonio NarinoCarrera 55 # 45-45BogotaCOLOMBIA

Treasurer: Emeritus Professor Peter TaylorPO Box 6165O’Connor ACT 2601AUSTRALIA

Regional Representatives

Africa: Professor John WebbDepartment of MathematicsUniversity of Cape TownRondebosch 7700SOUTH AFRICA

Asia: Mr Pak-Hong CheungMunsang College (Hong Kong Island)26 Tai On StreetSai Wan HoHong KongCHINA

Europe: Professor Nikolay KonstantinovPO Box 68Moscow 121108RUSSIA

Professor Francisco Bellot-RosadoRoyal Spanish Mathematical SocietyDos De Mayo 16-8#DCHAE-47004 ValladolidSPAIN

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North America: Professor Harold ReiterDepartment of MathematicsUniversity of North Carolina at Charlotte9201 University City Blvd.Charlotte, NC 28223-0001USA

Oceania: Professor Derek HoltonDepartment of Mathematics and StatisticsUniversity of OtagoPO Box 56DunedinNEW ZEALAND

South America: Professor Patricia FauringDepartment of MathematicsBuenos Aires UniversityBuenos AiresARGENTINA

The aims of the Federation are:

1. to promote excellence in, and research associated with,mathematics education through the use of school math-ematics competitions;

2. to promote meetings and conferences where persons inter-ested in mathematics contests can exchange and developideas for use in their countries;

3. to provide opportunities for the exchanging of informationfor mathematics education through published material, no-tably through the Journal of the Federation;

4. to recognize through the WFNMC Awards system personswho have made notable contributions to mathematics edu-cation through mathematical challenge around the world;

5. to organize assistance provided by countries with devel-oped systems for competitions for countries attempting todevelop competitions;

6. to promote mathematics and to encourage young mathe-maticians.

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From the President

Dear Fellow Federalists!

I hope you are enjoying happy and productive lives. The 7th Congress ofour World Federation of National Mathematics Competitions is beingorganized by our Past President Dr Maria de Losada. It will takeplace on July 21–24 (arrival on July 20) 2014 at the Hotel El Pradoin Barranquilla, Colombia, and will include a half-day excursion to thehistoric Cartagena. I hope to see you all there. Full details shouldappear in the next issue of the journal. So, plan your travel and startyour creative engines!

During our previous congresses and our sections at ICME, the ProgramCommittee was not always utilized. I know that first hand as a longterm member of the committee and its past chair. I hope this situationwill change, and Dr Kiril Bankov’s Program Committee will be activelyinvolved in creating academic structure of our 2014 Congress.

I believe that our 2010 Riga Congress had a sound basic structure, builtaround the following four sections:

1. Competitions around the World

2. Creating Competition Problems and Problem Solving

3. Work with Students and Teachers

4. Building Bridges between Research and Competition Problems

This can serve as a foundation for the 2014 Congress and beyond. Inaddition, we should have a number of plenary lectures, workshops, anddiscussions on issues important to us. Another session was successful inRiga: a session where we presented our favorite problems and solutions,which resulted in a small book. I hope such a session will take place in2014 and more participants will submit their problems, thus creating alarger book than the one published in Riga.

There is no time for complacency. We can and ought to pull our ef-forts together and sustain a vibrant World Federation. It means work,

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hard work, but work we love. As the brave American journalist Ed-ward R. Murrow (the one who in 1954 challenged U.S. Senator JosephMcCarthy on live television) said on October 15, 1958,

“Our history will be what we make it. If we go on as we are, thenhistory will take its revenge, and retribution will not limp in catchingup with us.”

Best wishes,

Alexander SoiferPresident of WFNMC

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From the Editor

Welcome to Mathematics Competitions Vol. 26, No. 1.

First of all I would like to thank again the Australian MathematicsTrust for continued support, without which each issue (note the newcover) of the journal could not be published, and in particular HeatherSommariva, Bernadette Webster and Pavel Calabek for their assistancein the preparation of this issue.

In July 2012 at the WFNMC miniconference in Seoul the new MC’sEditorial Board was formed. It comprises Waldemar Pompe (Poland),Sergey Dorichenko (Russia), Alexander Soifer and Don Barry (bothUSA).

Submission of articles:

The journal Mathematics Competitions is interested in receiving articlesdealing with mathematics competitions, not only at national and inter-national level, but also at regional and primary school level. There aremany readers in different countries interested in these different levels ofcompetitions.

• The journal traditionally contains many different kinds of arti-cles, including reports, analyses of competition problems and thepresentation of interesting mathematics arising from competitionproblems. Potential authors are encouraged to submit articles ofall kinds.

• To maintain and improve the quality of the journal and its use-fulness to those involved in mathematics competitions, all articlesare subject to review and comment by one or more competent ref-erees. The precise criteria used will depend on the type of article,but can be summarised by saying that an article accepted mustbe correct and appropriate, the content accurate and interesting,

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and, where the focus is mathematical, the mathematics fresh andwell presented. This editorial and refereeing process is designed tohelp improve those articles which deserve to be published.

At the outset, the most important thing is that if you have anythingto contribute on any aspect of mathematics competitions at any level,local, regional or national, we would welcome your contribution.

Articles should be submitted in English, with a black and white photo-graph and a short profile of the author. Alternatively, the article canbe submitted on an IBM PC compatible disk or a Macintosh disk. Weprefere LATEX or TEX format of contributions, but any text file will behelpful.

Articles, and correspondence, can also be forwarded to the editor by mailto

The Editor, Mathematics CompetitionsAustralian Mathematics TrustUniversity of Canberra Locked Bag 1Canberra GPO ACT 2601AUSTRALIA

or to

Dr Jaroslav SvrcekDept. of Algebra and GeometryPalacky University of Olomouc17. listopadu 1192/12771 46 OLOMOUCCZECH REPUBLIC

[email protected]

Jaroslav SvrcekJune 2013

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What “Problem Solving” Ought to Meanand how Combinatorial Geometry

Answers this Question:Divertismento in Nine Movements

Alexander Soifer

Born and educated in Moscow, Alexan-der Soifer has for 33 years been a Pro-fessor at the University of Colorado,teaching math, and art and film his-tory. He has published over 200 arti-cles, and a good number of books. Inthe past 3 years, 6 of his books have ap-peared in Springer: The MathematicalColoring Book: Mathematics of Color-ing and the Colorful Life of Its Cre-ators; Mathematics as Problem Solv-

ing: How Does One Cut a Triangle?; Geometric Etudes in CombinatorialMathematics; Ramsey Theory Yesterday, Today, and Tomorrow; and ColoradoMathematical Olympiad and Further Explorations. He has founded and for29 years ran the Colorado Mathematical Olympiad. Soifer has also servedon the Soviet Union Math Olympiad (1970–1973) and USA Math Olympiad(1996–2005).

The goal of mathematics education should be showing in a class-room what mathematics is and what mathematicians do. This canbe achieved not by teaching but rather by creating an atmosphere inwhich students learn mathematics by doing it. As in “real” math-ematics, this can be done by solving problems that require not justdeductive reasoning, but also experiments, construction of examples,and synthesis in a single problem of ideas from various branches ofmathematics. My recent five Springer books provide just right illus-trations of these ideas, fragments of “live” mathematics.

Give a man a fish, and you will feed him for a day.Teach a man how to fish, and you will feed him for a lifetime.

Laozı (VI century BC)

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1 Laozı and a Problem Solving Approach to Life

The great Chinese sage Laozı proposes to teach a man to fish as a methodof solving the problem of survival. This does go much further than givinga man a fish. However, is it good enough in today’s rapidly changingworld? We will come back to Laozı at the end of this essay.

2 The Principal Goal of Mathematics Education

What is the main goal of mathematics education?

Is it passing standardized three-letter tests, such as SAT, ACT, GRE,KGB, CIA (well, the two latter triples are from a different opera :-)).

Is it “teaching to the test,” as USA President George W. Bush believed?

Most would agree that problem solving is the goal. Fair enough, butthere is no agreement on the answer to a natural question: Just what isproblem solving?

A typical secondary school problem looks like A ⇒ B, i.e., given A proveB by using theorem C. In real life, no one gives a research mathematiciana B, it is discovered by intuition and is based on experimentation. And ofcourse, no one has a C since nobody knows what would work to solve theproblem which is not yet solved: a research mathematician is a pioneer,moving along an untraveled road!

And so, we ought to bring the school mathematics as close as possible toresearch mathematics. We ought to let our students experiment in ourclassroom-laboratory. We ought to let them develop intuition and use itto come up with conjectures B. And we ought to let our students findthat tool, theorem C that does the job of deductive proving.

In my opinion, the real goal of mathematical education is to demonstratein the classroom what mathematics is, and what mathematicians do.

3 Experiment in Mathematics

I have a good news and bad news for you. The bad news is, it is impos-sible to “teach” problem solving. The good news is, we can create an

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environment in which students learn problem solving by solving prob-lems on their own, with our gentle guidance. And I do not single outmathematics—all fields of human endeavour are about problem solving.Alright, but what kind of problems should we offer our students?

First of all, we ought to set up a mathematical laboratory, where stu-dents conduct mathematical experiments, develop inductive reasoningand create conjectures. The following classic example comes from [2].

Partitioning the Plane. In how many regions do n straight lines ingeneral position partition the plane?

We say that several straight lines on the plane are in general position ifno two lines are parallel and no three lines have a point in common.

Solution. Let us denote by S(n) the number of regions into whichn straight lines in general position partition the plane. Now let usexperiment: we draw one line on the plane and get S(1) = 2; we addanother line to see that S(2) = 4; we add one more line to shows thatS(3) = 7; and one more line demonstrates that S(4) = 11 (please seeFigures 1 and 2).

Figure 1. S(3) = 7 Figure 2. S(4) = 11

Let us put the data in a table:

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Number of Lines N S(n) Difference S(n)− S(n− 1)

1 2 2

2 4

3

3 7

4

4 11

We notice thatS(n) = S(n− 1) + n.

This recursive formula strikingly resembles the growth in the well-knownequality 1 + 2 + · · · + n = 1

2n(n + 1), i.e., if S1(n) = 1 + 2 + · · · + n,we would get the same recursive relationship as in our original problem:S1(n) = S1(n− 1) + n.

So let us check the hypothesis S(n) = 12n(n+ 1):

n S(n) 12n(n+ 1)

1 2 1

2 4 3

3 7 6

4 11 10

Our hypothesis does not work, but we can now see from the table abovethat S(n) and 1

2n(n + 1) differ only by 1, and apparently always by 1!Thus, we can now conjecture:

S(n) =1

2n(n+ 1) + 1.

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All there is left is to prove our conjecture, say, by mathematical induc-tion.

As we already know, our conjecture holds for n = 1.

Assume that it is true for n = k, i.e., k straight lines in general positionpartition the plane into S(k) = 1

2k(k + 1) + 1 regions.

Let now n = k + 1, i.e., let k + 1 straight lines in general positionbe given in the plane. If we remove one of the lines L, then by theinductive assumption the remaining k lines would partition the planeinto S(k) = 1

2k(k + 1) + 1 regions. Since we have k + 1 lines in generalposition, the remaining k lines all intersect the line L; moreover, theyintersect L in k different points a1, a2, . . . , ak (see Figure 3).

a1 a2 ak. . .

Figure 3

These k points split the line L into k+1 intervals. Each of these intervalssplits one region of the partition of the plane by k lines into two newregions, i.e. instead of k+1 old regions we get 2(k+1) new regions, i.e.,

S(k + l) = S(k) + (k + l),

therefore,

S(k + l) =1

2k(k + 1) + 1 + (k + l) =

1

2(k + 1)(k + 2) + 1.

In other words, our conjecture holds for n = k+1. Thus, n straight linesin general position partition the plane into 1

2n(n+ 1) + 1 regions.

4 Construction of Examples in Mathematics

Construction of examples and counterexamples plays a major role inmathematics, amounting to circa 50 % of its results. In fact, the greatRussian mathematician Israel M. Gelfand once said,

Theories come and go; examples live forever.

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Yet, practically the entire school mathematics consists of analyticalproofs. In order to bring instruction closer to “the real mix” we oughtto include construction of examples and counterexamples in education.Let me share one example, where a construction solves the problem [5].

Positive2 (A. Soifer, 2001). Is there a way to fill a 2001×2001 squaretable T with pluses and minuses, one sign per cell of T , such that noseries of interchanging all signs in any 1000×1000 or 1001×1001 squareof the table can fill T with all pluses?

Solution. Having created this problem and a solution for the 2001 Col-orado Mathematical Olympiad, I felt that another solution was possibleusing an invariant, but failed to find it. Two days after the Olympiad,on April 22, 2001, the past double-winner of the Olympiad Matt Kahle,now Professor at Ohio State University, found the solution that eludedme. It is concise and beautiful.

Define (please see Figure 4)

Φ = {the set of all cells of T , except those in the middle row}.

Figure 4

Observe that no matter where a 1000 × 1000 square S is placed in thetable T , it intersects Φ in an even number of cells, because there are 1000equal columns in S. Observe also that no matter where a 1001 × 1001square S′ is placed in T , it also intersects Φ in an even number of unit

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squares, because there are 1000 equal rows in S′ (one row is alwaysmissing, since the middle row is omitted in S.)

We can now create the required assignment of signs in T that cannotbe converted into all pluses. Let Φ have any assignment with an oddnumber of + signs, and the missing in Φ middle row be assigned signsin any way. No series of operations can change the parity of the numberof pluses in Φ, and thus no series of allowed operations can create allpluses in Φ.

5 Method & Anti-Method

Tiling with Dominoes. (Method). Can a chessboard with two diag-onally opposite squares missing, be tiled by dominoes (Figure 5)?

Figure 5

Solution. Color the board in a chessboard fashion (Figure 6). No matterwhere a domino is placed on the board, vertically or horizontally, itwould cover one black and one white square.

Figure 6

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Thus, it is necessary for tileability to have equal numbers of black andwhite squares in the board—but they are not equal in our truncatedboard. Therefore, the required tiling does not exist.

It is certainly impressive and unforgettable to see for the first time howcoloring can solve a mathematical problem, as we have witnessed here.However, I noticed that once a student learns this coloring idea, healways resorts to it when a chessboard and dominoes are present in theproblem. This is why I created the following “Anti-Method” Problemand used it in the Colorado Mathematical Olympiad [5].

The Tiling Game (Anti-Method, A. Soifer, 1989). Mark and Juliaare playing the following tiling game on a 1998 × 1999 board. They inturn are putting 1×1 square tiles on the board. After each of them madeexactly 100 moves (and thus they covered 200 squares of the board) awinner is determined as follows: Julia wins if the tiling of the board canbe completed with dominoes. Otherwise Mark wins. (Dominoes are 1×2rectangles, which cover exactly two squares of the board.) Can you finda strategy for one of the players allowing him to win regardless of whatthe moves of the other player may be? You cannot? Let me help you:Mark goes first!

Solution. Julia (i.e., the second player) has a strategy that allows herto win regardless of what Mark’s moves may be. All she needs is abit of home preparation: Julia prepares a tiling template showing oneparticular way, call it T , of tiling the whole 1998 × 1999 chessboard.Figure 7 shows one such tiling template T for a 8× 13 chessboard.

The strategy for Julia is now clear. As soon as Mark puts a 1 × 1 tileM on the board, Julia puts her template T on the board to determinewhich domino of the template T contains Mark’s tile M . Then she putsher 1 × 1 tile J to cover the second square of the same domino (Figure8).

6 Synthesis & Combinatorial Geometry

School mathematics consists predominantly of problems with single-ideasolutions, found by analysis. We ought to introduce a sense ofmathemat-

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8

13

Figure 7. Tiling template for a 8× 13 chessboard

8

13

M

J

Figure 8. Winning Strategy

ical reality in the classroom by presenting synthesis, by offering problemsthat require for their solution ideas from a number of mathematical dis-ciplines: geometry, algebra, number theory, trigonometry, linear algebra,etc.

And here comes Combinatorial Geometry! It offers an abundance ofproblems that sound like a “regular” secondary school geometry, butrequire for their solutions synthesis of ideas from geometry, algebranumber theory, trigonometry, ideas of analysis, etc. See for example[3]; [4]; and [1].

Moreover, combinatorial geometry offers us open-ended problems. Andit offers problems that any geometry student can understand, and yet no

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one has yet solved! Let us stop this discrimination of our students basedon their young age, and allow them to touch and smell, and work on realmathematics and its unsolved problems. They may find a partial advanceinto solutions; they may settle some open problems completely. And theywill then know the answer to what ought to become the fundamentalquestions of mathematical education:

What is Mathematics?What do mathematicians do?

In fact, I would opine that every discipline is about problem solving.And so the main goal of every discipline ought to be to enable studentsto learn how to think within the discipline, how to solve problems of thediscipline, and finally what that discipline is, and what the professionalswithin the discipline do!

And more generally, we can ask, what is life about? I believe that

Life itself is about overcoming difficulties, i.e., solving problems.

And mathematics to sciences does what gymnastics does to sports:

Mathematics is gymnastics of the mind.

Doing mathematics develops a universal approach to problem solvingand intuition that go a long way in preparing our students for solvingproblems, no matter what kind of problems they will face in their lives.

7 Open Ended and Open Problems

As a junior at the university, I approached my supervisor ProfessorLeonid Yakovlevich Kulikov with an open problem I liked (he was mysupervisor ever since my freshman year). He replied, “Learn first, thetime will come later to enter research.” He meant well, but politicallyspeaking, this was discrimination based on my age. Seeing my disap-pointment, Kulikov continued, “It does not look like I can stop you

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from doing research. Alright, whatever results you obtain on this openproblem, I will count them as your course work.” Soon I received firstresearch results, and my life in mathematics began.

We ought to allow our students to learn what mathematicians do byoffering them not just unrelated to each other exercises but rather seriesof problems leading to a deeper and deeper understanding. And weought to let students “touch” unsolved problems of mathematics, givethem a taste of the unknown, a taste of adventure and discovery. Andcombinatorial geometry serves these goals very well, providing us witheasy-to-understand, hard-to-solve—or even unsolved—problems!

I will formulate here two examples. The volume of this essay does notallow including the solutions. You can find them in my five recentSpringer books, listed in References.

Points in a Triangle [3] Out of any n points in or on the boundaryof a triangle of area 1, there are 3 points that form a triangle of area atleast 1

4 .

a) Prove this statement for n = 9.

b) Prove this statement for n = 7.

c) Prove this statement for n = 5.

d) Show that the statement is not true for n = 4, thus making n = 5best possible.

Chromatic Number of the Plane [1] No matter how the plane iscolored in n colors, there are two points of the same color distance 1apart.

a) Prove this statement for n = 2.

b) Prove this statement for n = 3.

c) Disprove this statement for n = 7.

d) The answer for n = 4, 5, and 6 is unknown to man—this is aforefront of mathematics!

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8 Beauty in Mathematics

I suspect you all have encountered examples of beauty of the mathe-matical kind. And so, I will share just one example, a problem thatmy high school friend and I created when we were fellow graduate stu-dents in Moscow. It was published in Kvant magazine, and we were evenhandsomely compensated!

Forty-One Rooks (A. Soifer and S. Slobodnik, 1973). Forty-onerooks are placed on a 10×10 chessboard. Prove that you can choose fiveof them that do not attack each other. (We say that two rooks “attack”each other if they are in the same row or column of the chessboard.)

Solution, first found during the 1984 Olympiad (!) by Russel Shaffer;the idea of using symmetry of all colors by gluing a cylinder, belongs tomy university student Bob Wood.

Let us make a cylinder out of the chessboard by gluing together twoopposite sides of the board, and color the cylinder diagonally in 10 colors(Figure 9).

Figure 9. One out of the ten one-color diagonals is shown

Now we have 41 = 4×10+1 pigeons (rooks) in 10 pigeonholes (one-colordiagonals), therefore, by the Pigeonhole Principle, there is at least one

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hole that contains at least 5 pigeons. But the 5 rooks located on thesame one-color diagonal do not attack each other!

9 Returning to Laozı 2600 Years Later

On September 5, 2011, I wrote a letter to the Editor of The New YorkTimes precisely on the debate taking place between two most popularapproaches to mathematical education. Let me present this letter herein its entirety:

In “In Classroom of Future, Stagnant Scores” (September 4, 2011,p. 1), Matt Richtel discusses two approaches to mathematical edu-cation, “embrace of technology” vs. “back to the basics.” Shouldwe not first ask, what are the goals of education? To paraphrasethe great Chinese thinker Laozı, I would say: Give student skills,and you will feed him in the short run. Let student learn ideas, solveproblems, and you will feed him for a lifetime.

In the new “embrace of technology” approach I support takingteacher off the pedestal of a lecturer: one cannot teach mathematics,or anything for that matter. Students can learn math, and anythingelse, only by doing it, with a gentle guidance of the teacher. Tech-nology in the classroom more often than not treats students likerobots, and preprograms them with skills of today. But technologynowadays changes rapidly, as does the societal demands for particu-lar skills. A student who has learned critical thinking and problemsolving will easier adopt to the rapidly changing world.

“Back to the basics” is also not the best solution, for it empha-sizes mind numbing drill, and it also treats students as robots andpreprograms them with skills.

The goal in the mathematical classroom ought to be a practicaldemonstration of what mathematics is and what mathematiciansdo. Everything, life herself included, is about overcoming difficulties,solving problems.

We ought to abandon standardized multiple choice testing of skills.There are more important things to test. Over the past 29 years, inthe Colorado Mathematical Olympiad, we offered middle and highschool students 5 problems and 4 hours to think and solve. We “test”not topics, not skills, but creativity and originality of thought.

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Providing public education is not just an ethical thing to do—it is aprofitable investment. Are there many jobs today for computer-illiteratepersons? And yet just one generation ago, computers were a monopolyof researchers, and one generation before that did not exist at all.

Not every education is as good an investment as another. And we oughtto go beyond Laozı and his famous lines:

Give a man a fish, and you will feed him for a day.Teach a man how to fish, and you will feed him for a lifetime.

Not quite, dear Sage. Not in today’s day and age. What if there is nomore fish? What if the pond has dried out? And your man has only oneskill, to fish?

Education ought to introduce students to ideas and how to solve prob-lems no matter what field. A problem solver will not die if the fish disap-pears in a pond—he’ll learn to hunt, to grow veggies, to solve whateverproblems life puts in his path.

And so, we will go a long way by putting emphasis not on training skillsbut on creating atmosphere for developing problem solving abilities andattitudes:

Enable a man to learn how to solve problems,And you will feed him for a lifetime!

References

[1] Soifer, A. (2009-1). The Mathematical Coloring Book: Mathematicsof Coloring and the Colorful Life of Its Creators, Springer, New York.

[2] Soifer, A. (2009-2). Mathematics as Problem Solving, Springer, NewYork.

[3] Soifer, A. (2009-3). How Does One Cut a Triangle?, Springer, NewYork.

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[4] Soifer, A. (2011-1).Geometric Etudes in Combinatorial Mathematics,Springer, New York.

[5] Soifer, A. (2011-2). Colorado Mathematical Olympiad and FurtherExplorations: From the Mountains of Colorado to the Peaks of Math-ematics, Springer, New York.

Alexander Soifer

University of Colorado at Colorado Springs

P. O. Box 7150, Colorado Springs, CO 80933

USA

E-mail: [email protected]

http://www.uccs.edu/˜asoifer/

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Mathematics Competitions and GenderIssues: a Case of the Virtual Marathon

Mark Applebaum & Margo Kondratieva & Viktor Freiman

Mark Applebaum (Ph.D.), Head ofTeachers disciplinary development inKaye Academic College of Education,Be’er-Sheva, Israel. He has a reachexperience in preparing of Mathemat-ics’ teachers for primary and secondaryschools. His interests cover the do-mains of Mathematical Competitions,Development of Critical and CreativeThinking and Teaching Mathematicsfor Gifted and Promising students.

He is author of more than 50 publi-cations, including 30 articles and 10books.

Dr. Margo Kondratieva received a PhDdegree in mathematics in 1994. Sheis an Associate professor at MemorialUniversity in Canada. Margo Kon-dratieva has about 16 years of teach-ing experience in undergraduate math-ematics. Her current research inter-ests in the area of mathematics edu-cation are development of mathemati-cal thinking and learning, and designof educational activities enriching stu-dents’ experiences within the subject.She organizes various types of math-ematics competitions for grade schoolstudents. Margo has about 25 publica-tions in peer-reviewed journals in theareas of mathematics and mathemat-ics education. She has co-authored 4books.

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Viktor Freiman is a full professor of theUniversite de Moncton, Canada. Hisresearch is focused on mathematicalchallenge, giftedness and use of tech-nology in the interdisciplinary andproblem-based learning perspective,authoring and co-authoring severalbooks, book chapters, and journalarticles. He was a co-chair on theTopic Study Group on Activities andPrograms for Gifted at the ICME-11in Monterrey, in 2008. He was a lo-cal organizer of the International Sym-posium on Mathematics and its Con-nections with the Arts and Sciences in2009. Since 2011, he is co-editor of thenew Springer Book Series on Mathe-matics Education in the Digital Era.

1 Introduction

In today’s technologically enhanced world mathematics competitions be-come available to more students who are interested in challenging tasks.In this paper we look at gender-related issues pertinent to participationin the Virtual Mathematical Marathon over two years. Our study con-centrates on the following questions: what were boys’ and girls’ partici-pation patterns and how successful were they in online problem-solvingcompetition.

The Virtual Mathematical Marathon (VMM, http://www8.umoncton.ca/umcm-mmv/index.php) is an online competition open to everybodywho is interested in solving challenging problems over a long period oftime. As an extension of the virtual interactive learning communityCAMI (Chantier d’Apprentissages Mathematiques Interactifs; Freimanet al., 2009), the Marathon provides Grade 3–9 students with challengingmathematical problems.

Besides CAMI’s regular Problem of the week activity that was conductedover the school year, we developed a long-term summer competition foryoung students who may have interest in solving more challenging tasks

24


in a form of competition. A new section became available in summer2008 and since that time, four summer rounds have been organized.Moreover, support received from the Canadian Natural Sciences andEngineering Research Council (Promoscience Grant, 2009–2011) helpedus to develop a bilingual version of the Marathon (French and English)and introduce a winter round since 2010, thus making it a year-roundcompetition. First results of the project based on 2008-2009 participa-tion data were presented at the PME-36 Research Forum (Freiman &Applebaum, 2009) and in a journal article (Freiman &Applebaum, 2011).This paper is based on the work presented in the Topics Study Group 34on mathematics competitions at the ICME-12 congress while extendingour investigation of gender-related issues in virtual environments.

2 Gender-related data on maths competitions: isthere an issue?

Several educators express a concern regarding gender difference in math-ematics performance and the underrepresentation of women in science,technology, engineering and mathematics (STEM) careers (NationalAcademy of Science, Beyond Bias and Barriers: Finding the potentialof women in academic science and engineering, 2006; Hyde et al., 2008).Gender inequity is particularly evident in data related to the number ofgirls participating in the International Math Olympiad, or the numberof female professors in university mathematics and engineering depart-ments (Hyde & Mertz, 2009). There are several ways in which thisproblem may be addressed.

First, psychologists are looking for gender differences in brain struc-ture, in hormones, in the use of brain hemispheres, nuances of cognitiveor behavioural development and consequent spatial and numerical abil-ities that may predispose males to a greater aptitude and success inmathematics (Halpern, 1997; Moir & Jessel, 1989). However, severalfindings reported in the literature regarding this matter are not consis-tent (Spelke, 2005), partly due to the fact that experience alters brainstructures and functioning (Halpern, et al., 2007).

Second, detailed measurements of students’ achievements in mathemat-ics are being performed by educators at different stages of schooling in

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an attempt to identify the moment of occurrence and further dynamicsof gender gaps in mathematics. Many studies are consistent in their ob-servation that the gender gap becomes more evident as students progresstowards higher grades, especially if testing involves advanced topics inmathematics and higher cognitive level items. In contrast to earlier find-ings, some more current data provide no evidence of a gender differencefavouring males emerging in the high school years (Hyde et al., 2008).

Yet another interesting observation is that “achievement gains are in-sufficient unless the self-beliefs of girls have changed correspondingly”(Lloyd, Walsh & Yailagh, 2005, p. 385). Research that views genderdifferences through the lenses of the attribution theory (see e.g. Ban-dura, 1997) suggests that girls tend to attribute their math successes toexternal factors and to effort and their failures to their own lack of abil-ity (self defeating pattern), whereas boys tend to attribute the causesof their successes to internal factors and their failure to external factors(self-enhancing pattern). Since it is better for an individual to attributesuccess to ability, rather than to effort, because ability attributions aremore strongly related to motivation and skill development (Schunk &Gunn, 1986), these patterns have explained in part girls’ poorer achieve-ment(Lloyd et al., 2005).

A report of the American Association of University Women How SchoolsShortchange Girls (1992) focused on girls being discouraged from study-ing math and science. The report indicates that “girls receive less atten-tion in the classroom than boys and less encouragement for their efforts.In addition, the study showed that many classrooms created the atmo-sphere of competition among students. Such an atmosphere played tothe strength of boys, who were socialized to compete, but often intimi-dated girls, who were more often socialized to collaborate.” (Williams,2006, p. 301)

A third way of addressing the gender gap in mathematics is to investigatethe influence of socio-cultural factors. According to Von Glaserfeld(1989), the context in which learners find themselves is important inthe acquisition of knowledge. First, it was found that parents havegreater expectations for sons regarding their mathematical performancethat they have for daughters, and this has an influence on the students’results (Leder, 1993). It was also observed that even talented and

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motivated girls “are not immune to the ill effects of gender bias” (Leedy,LaLonde, & Runk, 2003, p. 290). In this respect it is unfortunate thatstereotypes that girls and women lack mathematical ability persist andare widely held by parents and teachers (Hyde et al., 2008). Leedy etal. (2003) studied beliefs held by students participating in regional mathcompetitions, as well as by their parents and teachers. They found thatmathematics is still viewed as a male domain by men, while girls andwomen fail to acknowledge the existence of the bias. They argue that thetask of the school is not to ignore or deny differences in learning styles,attitudes and performance but to acknowledge them and use them todevelop strategies aimed at providing gender equitable education.

In conclusion, in all three perspectives in research on gender in mathe-matics—cognitive, instructional, and socio-cultural—care is needed inconsidering how the data are collected, examined and interpreted be-cause within no approach is there a fully consistent theory that couldexplain the existing gender difference observed at the higher level ofmathematical tasks. As Halpern et al. (2007) point out, “there are nosingle or simple answers to the complex question about sex difference inmathematics”, and all “early experience, biological factors, educationalpolicy, and cultural context” need to be considered when approachingthis question.

3 Technology and gender: what patterns emerge inmathematics competitions?

While the previous section summarizes research related to gender issuesin mathematics education showing no conclusive findings, similar obser-vations can be drawn from technology-related studies that we will reviewvery briefly. For instance Fogasz (2006) reports that when talking aboutclassroom practices that involve computers as a learning tool, mathe-matics teachers held gender-based beliefs about their students that theincorporation of technology has more positive effects on males’ classroomengagement and on their affective responses, and thus using a techno-logical approach benefits boys’ learning to a greater extent.

At the same time Wood, Viskic & Petocz (2003) found no gender differ-ences in the students’ use of computers or in their attitudes towards the

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use of computers. This agrees with ideas expressed by Willams (2006)quoted above, who reviews studies showing that girls are as confident andactive as are boys in creating webpages, writing blogs, reading websites,and chatting online, among other activities.

As we mentioned in our earlier publications (Freiman et al., 2009;Freiman & Applebaim, 2009), the internet can be a suitable challengingenvironment for organizing mathematics competitions and problem solv-ing activities, contributing potentially to the development of mathemat-ical ability and giftedness. In a recent analysis of middle-school studentsparticipating in a web-based mathematics competition Carreira et al.(2012) argue that although it cannot be said that by solving problemsonline, students do better in mathematics, their data provide us withan evidence that the use of technology tends to involve more complexmathematical thinking.

Moreover, the use of technology can be considered as an inclusive formof mathematical enrichment, providing a tool, an inspiration, or a po-tentially challenging and motivating independent learning environmentfor any student. For the gifted ones, it is often a means to reach theappropriate depth and breadth of curriculum, to advance at the appro-priate pace for each learner, as well as to achieve better engagementand task commitment (Johnson, 2000; Jones & Simons, 2000; Renninger& Shumar, 2004; Freiman & Lirette-Pitre, 2009; Sullenger & Freiman,2011).

Being a part of a powerful set of out-of-regular-classroom activities suchas mathematical clubs, mathematical camps, mathematics competitions(Olympiads), on-line mathematics competitions play a significant role innurturing interest and motivating young learners of mathematics, as wellas in identification and fostering the most able and talented (Skvortsov,1978; Karnes & Riley, 1996; Robertson, 2007; Bicknell, 2008). Thechoice of appropriate challenging tasks is also an important condition ofthe success of mathematics competitions in developing students’ learn-ing potential. Leikin (2004, 2007) claims such tasks must be appropriateto students’ abilities, neither too easy, nor too difficult. They shouldmotivate students to persevere with task completion and develop math-ematical curiosity and interest in the subject. As well, they must supportand advance students’ beliefs about the creative nature of mathematics,

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the constructive nature of the learning process, and the dynamic na-ture of mathematical problems as having different solution paths andsupporting individual learning styles and knowledge construction.

While designing our Virtual Mathematical Marathon, we aimed to pro-vide students with an opportunity to discover their talent, which theycannot normally demonstrate in the regular classroom (Taylor, Gour-deau & Kenderov, 2004) thus we considered the marathon as a stimulusfor improving students’ informal learning. Fomin, Genkin & Itenberg(2000) described that during the marathon that they conducted on aface-to-face basis, their students managed to increase the number ofproblems they solved, relatively to other non-competing frameworks inwhich the same students participated.

Regarding gender issues in virtual mathematics competitions, we founda lack of data that we aim to address in our paper. In the followingsection we describe the Virtual Mathematical Marathon’s structure thatallowed us to collect data about participants, including data accordingto their gender. The main question we asked in our study was: Whatkind of differences have been observed in boys’ and girls’ behaviour duringtheir participation in VMM? We divided our investigation in two partsaddressing the following two sub-questions:

– Was there a gender difference in the initial enrolment of student-participants of VMM? How did participation evolve during the com-petition, according to the gender?

– What were the gender-related patterns in participants’ behaviouraccording to the difficulty levels for each year in terms of both,participation and success rate?

4 Structure of the virtual mathematical marathon

According to our model of the VMM, one set of 4 non-routine chal-lenging problems was posted twice a week on the CAMI website (www.umoncton.ca/cami) over 10 weeks, from June to August in 2008 and2009. In total, 20 sets of problems were offered to the participants ofeach round. Every registered member could login, choose a problem,solve it, and submit an answer by selecting it from a multiple-choicemenu. The automatic scoring system immediately evaluated students’

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success producing a score for the problems and adjusting a total scorethat affected the overall standing.

According to the level of difficulty, scores per problem were determinedas follows: level 1(easiest) was scored with 3 points, level 2 with 5 points,level 3 with 7 points, and level 4 (hardest) with 10 points. To supportstudents’ participation in the marathon, unsuccessful attempts were alsorewarded with 1, 2, 3, and 4 points respectively. Participants could jointhe marathon, solve as many problems as they wished, withdraw, andcome back at any time. The tasks were developed by a team of expertsin mathematics and mathematics education.

Here are examples of such tasks coming from one set:

Level 1 problem: How many numbers from 10 to 200 have the prop-erty that reversing their digits does not change the number?

A) 17 B) 18 C) 19 D) 20

Comment: students can approach this problem by simply listing allnumbers with the required property. These numbers are 11, 22, 33,44, 55, 66, 77, 88, 99, 101, 111, 121, 131, 141, 151, 161, 171, 181, and191. Thus the answer is 19.

Level 2 problem: Two dice are thrown at random. What is theprobability that the two numbers shown are the digits of a two-digitperfect square?

A) 1/9 B) 5/36 C) 2/9 D) 5/18

Comment: students need to be familiar with the notion of probability,some counting techniques, and apply reasoning. They should notice thatthe only 2-digit perfect squares that can be constructed from digits 1,2, 3, 4, 5, 6, are 16, 25, 36, and 64. This gives 4 possible squares, withtwo ways for each to occur. Since there are 36 possible outcomes, theprobability is 8/36 = 2/9.

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Level 3 problem: The volume of a cube is 64 units cubed. What isthe surface area of the cube?

A) 16 units squared B) 64 units squaredC) 96 units squared D) 256 units squared

Comment: students need to know basic facts about cubes. They couldreason as follow. If the total volume is 64, then the side length is 4units. This means each face has area 4 × 4 = 16. There are 6 faces, sothe surface area is 6× 16 = 96 units squared.

Level 4 problem: You have 100 tiles, numbered 0 to 99. Take any setof three tiles. If the number on one of the tiles is the sum of the othertwo numbers, call the set “good”. Otherwise, call the set “bad”. Howmany good sets of three tiles are there?

A) 160 B) 1600 C) 1225 D) 2401

Comment: students need to notice a pattern and invent some usefulcounting technique in order to solve this problem. For example, theymay reason as follows. If 3 is the largest number, there is one good set,{1, 2, 3}. If 4 is the largest, there is one good set, {1, 3, 4}. For 5 and 6there are two, {1, 4, 5}, {2, 3, 5} and {1, 5, 6}, {2, 4, 6}, respectively. For7 and 8 there are three each, for 9 and 10 there are four each, and so on.In this way, when we reach 97 there are 48 good sets, for 98 there arealso 48, and for 99 there are 49. It remains to compute the sum

1 + 1 + 2 + 2 + 3 + 3 + · · ·+ 47 + 47 + 48 + 48 + 49

= 49 + (1 + 48) + (1 + 48) + (2 + 47) + (2 + 47)

+ · · ·+ (24 + 25) + (24 + 25) = 49× 49 = 2401.

5 Proving and disproving conjectures: fosteringexploration and questioning

The participants of the marathon were all members of the CAMI commu-nity. They received an invitation by email to take part in the marathon.Most of them were from New Brunswick, Canada. We also had a few

31


participants from Quebec and from France. We have no reliable data onstudents’ age, but the most frequent CAMI users are Grades 6–8 (ages12–14) which is a good approximation.

In order to investigate the first sub-question, we have collected andanalysed the data about boys’ and girls’ participation for each of 20sets (in both years). We collected and compared the numbers of initialenrolment and on-going visits for boys and girls separately.

In order to address the second sub-question, we have analysed the dataabout boys’ and girls’ attempts to solve either all or some particularproblems from each set. For example, some students could attemptonly easier questions (levels 1–2). We were interested to see if thestudent was trying to stay in a ‘safer’ zone, or to take some greater‘risks’ solving more challenging problems (levels 3–4). In this respect, wewere curious whether a virtual problem-solving environment had allowedgirls to exhibit risk-taking behaviour at a rate comparable to the oneof boys. Moreover, we draw on our data from previous analyses thatemphasized particular behaviour of students who were the most activeand successful (the winners of each 20-round game), the group we calledthe ‘most persistent’(Freiman & Applebaum, 2011). We have comparedthe number of girls and boys among this group. The next section presentsour findings.

6 Preliminary results and discussion

There were 298 students (194 in the first year and 104 in the secondyear) who participated in at least one round (of the total of 20 roundseach year) of the marathon. In the first year, there were more boys(n = 110, or 56.7 %) than girls (n = 84, or 43.3 %). In the second yearthe number of girls was slightly higher than number of boys (n = 56,or 53.8 % against n = 48, or 46.2 %). Over two years, our data didnot indicate any significant difference in participation according to thegender: girls seem to be as active as boys.

Further, Figures 1 and 2 below show how the number was changing overeach competition. From Figure 1, we learn that in the first three setsof the Year 1, the number of boys was higher than number of girls,but starting from set 7, the numbers are nearly the same in each of

32


remaining sets. We can see therefore that girls who decided to continuetheir participation were as persistent as boys. A similar trend can beobserved in Year 2 data (Figure 2); while the number of participants ismuch lower than in the first year, there were still more boys than girlsin the first set but starting from the set 7, the number of girls and boyswas nearly the same until the end of the competition. It is remarkablethat among the winners of the two 20-round games there were 6 boysand 5 girls, so nearly the same number of each gender.

The repartition of the number of attempts by gender, according to Table1 (Year 1 and 2) shows that there was no significant difference in thenumber of attempts related to the difficulty levels between girls andboys. Usually, participants who tried a problem of level 1 (easiest) didattempt to solve problems of other levels; some difference is only betweenthe levels one and two for both genders. This observation is particularlyvaluable in view of the fact that in a regular classroom setting “teachersperceived that girls. . . produced fewer exceptional, risk-taking [learners]than did boys.” (Williams, 2006).

The dynamics of success rates is similar between the girls and the boysin the first year; also, both genders were more successful on easier levels(1 and 2) and less in more difficult levels (3 and 4). In the Year 2,however, the boys have clearly outperformed girls at all levels; with thesame trends between levels 1–2 (easier—better solved) and 3–4 (harder—less success).

7 Conclusive remarks

The paper explores gender-related data of students’ participation in avirtual mathematics competition, a marathon. Based on studies thatindicate virtual environments have the potential to attract as manygirls as boys to take part in solving challenging problems, we analysedparticipation patterns and success rate at the VMM, according to thegender. While observing participation over a long period of time duringthe first two years of the competition, we found that for both years girlsand boys showed similar patterns when, after first few weeks, a number ofparticipants who decided to stay remains relatively stable, independentlyof success or failure on certain tasks, thus demonstrating risk-taking

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behaviour and persistence for both genders. This similarity betweenboth genders is consistent with other researchers’ findings (Lloyd et al.,2005; Williams, 2006) indicating non-significant gender difference at thejunior high level in mathematics as well as equal abilities and interest ofboth boys and girls to participate in online activities.

Our preliminary data analysis has several limitations: the major beingthat we neither analysed students’ solutions nor conducted interviewsasking them about their level of satisfaction with the game. However,from the data of the online survey, we learn that both, girls and boys,seem to express their interest for more challenging mathematics and toappreciate the experience.

When asked “What motivates you to take part in Marathon?”, studentsanswered: “desire to improve my mathematical abilities”, “I love math”,“. . . it [VMM] allows me to practice my mathematics”, “I like the factthat lot of students compete and the questions come with solutions withdetailed explanations in case of a mistake”, “I like to see how high Ican rank with everybody else. I especially like to getting top 10 andgetting name on the home page”, “I love to solve math problems”, “Ilike math and solving problems. Plus I like to compete”, “Challenge bynew, interesting, unusual problems”.

For our question “What do you like about Virtual Mathematical Mara-thon?”, students’ answers were: “I get to do challenging problems. Theexplanations are good.”, “It’s fun and helps me learn more about math”,“Short problems, ranking system, solutions are available immediately”,“It’s possible to do it anytime during the week, so I can easily workaround homework”, “That it makes me think and sweat”, “It challengesme and I love it all. I like seeing the problems every week”. Suchstudents’ responses are very encouraging and reassuring that the goal ofbuilding a virtual community of junior mathematics problem solvers setby our team can be achieved by continuing our collaborative work onthe VVM project.

Our future work will use more data and look at more detailed data anal-ysis including students’ interviews that could allow for a deeper insightinto students’ behaviour and better understanding of their thoughts andattitudes about the online problem-solving activity.

34


Figures 1 and 2: Dynamics of students’ participation: total (3), boys(2), and girls (1), X - Set number; Y - Number of participants

35


Le

vel 1

Le

vel 2

Le

vel 3

Le

vel 4

To

tal

First year B

oys

Num

ber o

f suc

cess

ful s

olut

ions

/

Num

ber o

f atte

mpt

s 21

112

4

202

115

20

189

20

072

81

440

0

Per c

ents

58

.77%

56

.93%

44

.28%

36

%

49.1

4%

Girl

s N

umbe

r of s

ucce

ssfu

l sol

utio

ns /

Num

ber o

f atte

mpt

s 13

781

12

170

12

260

12

239

50

225

0

Per c

ents

59

.12%

57

.85%

49

.18%

31

.97%

49

.80%

Second year

Boy

s N

umbe

r of s

ucce

ssfu

l sol

utio

ns /

Num

ber o

f atte

mpt

s 11

978

10

865

10

551

10

450

43

624

4

Per c

ents

65

.55%

60

.19%

48

.57%

48

.08%

55

.96%

Girl

s N

umbe

r of s

ucce

ssfu

l sol

utio

ns /

Num

ber o

f atte

mpt

s 10

652

9445

9829

10

034

39

816

0

Per c

ents

49

.06%

47

.87%

29

.59%

34

%

40.2

0%

Table 1: The boys’ and girls’ success rates at each difficulty level

36


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[8] Freiman, V., & Applebaum, M. (2011). Online Mathematical Com-petition: Using Virtual Marathon to Challenge Promising Studentsand to Develop Their Persistence. Canadian Journal of Science,Mathematics and Technology Education, 11(1), 55–66.

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[19] Leikin, R. (2007). Habits of mind associated with advanced math-ematical thinking and solution spaces of mathematical tasks. TheFifth Conference of the European Society for Research in Mathe-matics Education—CERME-5. (pp. 2330–2339) (CD-ROM and On-line). http://ermeweb.free.fr/Cerme5.pdf Retrieved September22, 2012.

[20] Lloyd, J. E. V., Walsh, J., & Yailagh, M. S. (2005). Sex differencesin performance attributions, self-efficacy and achievement in math-ematics: if I’m so smart, why don’t I know it? Canadian Journalof Education, 28(3), 384–408.

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handle/10090/13996/1087.pdf?sequence=1 Retrieved April 15,2010.

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[28] Spelke, E. S. (2005). Sex differences in intrinsic aptitude for math-ematics and science? American Psychologist, 60, 950–958.

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8 Acknowledgments

We thank the National Science and Engineering Research Council ofCanada (Promoscience Grant), Canadian Mathematical Society, New-Brunswick Innovation Foundation, and Ministere de l’Education et duDeveloppement de la petite enfance du Nouveau-Brunswick for theirsupport on the development of the Virtual Mathematical Marathon.

Mark ApplebaumKaye Academic College of [email protected]

Margo KondratievaMemorial [email protected]

Viktor FreimanUniversity de [email protected]

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The Good, the Bad and the Pleasure (notPressure!) of Mathematics Competitions

Siu Man Keung

Siu Man Keung obtained a BSc asa double major in mathematics andphysics from University of Hong Kongand went on to earn a PhD in mathe-matics from Columbia University. Likethe Oxford cleric in Chaucer’s TheCanterbury Tales, “and gladly wouldhe learn, and gladly teach”. He hadbeen doing that for more than threedecades until he retired in 2005. Hehas published some research papersin mathematics and computer sci-ence, some more papers in history ofmathematics and mathematics educa-tion, and several books in populariz-ing mathematics. In particular he ismost interested in integrating historyof mathematics with the learning andteaching of mathematics, and has beenparticipating actively in an interna-tional community of History and Ped-agogy of Mathematics since the mid1980s.

1 Introduction

A mathematics competition, as but one among many different kindsof extracurricular activity, should enhance the teaching and learning ofmathematics in a positive way rather than present a controversy in anegative way. But why does it sometimes become a controversial issuefor some people as to its negative effect? What are the pros and cons ofthis extracurricular activity known as a mathematics competition?

I cannot claim myself to be actively involved in mathematics competi-tions, but will attempt to share some of my views of this activity gleaned

41


from the limited experience gained in the past years. The intention isto invite more discussion of the topic from those who are much moreexperienced and know much more about mathematics competitions1.

To begin with I’d like to state clearly which aspect of this activityI will not touch on. In the recent decade mathematics competitionshave mushroomed into an industry, some of which are connected withprofit-making or fame-gaining intention. Even if these activities mayhave indirect benefit to the learning of mathematics, which I seriouslydoubt, an academic discussion of the phenomenon is, mathematicallyspeaking, irrelevant. Rather, it is more a topic of discussion for its socialand cultural aspect, namely, what makes parents push their children tothese competitions and to training centres which are set up to preparethe children for these competitions, sometimes even against the likingof the children? With such a disclaimer let me get back to issueson mathematics competitions that have to do with mathematics andmathematics education.

The “good” of mathematics competitions

The only experiences I had of an international mathematics competitionoccurred in 1988 and in 1994. I helped as a coach when Hong Kong firstentered the 29th IMO (International Mathematical Olympiad) in 1988held in Canberra, and I worked as a coordinator to grade the answerscripts of contestants in the 35th IMO in 1994 held in Hong Kong.Through working in these two instances I began to see how the IMOcan exert good influence on the educational side, which I had overlookedbefore. I wrote up my reflections on the IMO in an article, from which Inow extract the three points on “the good” of mathematics competitions[4, pp. 74–76].

1. All contestants know that clear and logical presentation is a neces-sary condition for a high score. When I read their answers I hada markedly different feeling from that I have in reading the answerscripts of many of my students. I felt cozy. Even for incomplete or

1This paper is an extended text of an invited talk given at a mathematics educationforum held in conjunction with the International Mathematics Competition scheduledon July 24–27, 2012 in Taipei. I thank the organizers, particularly SUN Wen-Hsien ofthe Chiu Chang Mathematics Education Foundation, for inviting me to give a talk sothat I can take the opportunity to organize my thoughts and share them with thosewho are interested in mathematics competitions.

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incorrect answers I could still see where the writing is leading me.But in the case of many of my students, despite advice, coaxing,plea, protest (anything short of threat!) on my part, they writedown anything that comes to their minds, disconnected, disorga-nized and perhaps irrelevant pieces. One possible reason for thisbad habit is the examination strategy they have adopted since theirschool days—write down everything you can remember, for you willscore certain marks for certain key points (even if these key pointsare not necessarily presented in a correct logical order!) and thekind-hearted examiner will take the trouble to sift the wheat fromthe chaff! Many undergraduates still follow this strategy. Correct orincorrect answer aside, the least we can ask of our students shouldbe clear communication in mathematics (but sadly we cannot).

2. All contestants know that one can afford to spend up to one and ahalf hour on each problem on average, and hence nobody expects tosolve a problem in a matter of minutes. As a result, most contestantspossess the tenacity and the assiduity required for problem solving.They will not give up easily, but will try all ways and means toprobe the problem, to view it from different angles, and to explorethrough particular examples or experimental data. On the contrary,many of my students, again too much conditioned by examinationtechniques since their school days, would abandon a problem oncethey discover that it cannot be disposed of readily by routine means.In an examination when one races against time, this technique mayhave its excuse. Unfortunately, many students bring the same habitinto their daily study. Any problem that cannot be disposed of in 3minutes is a difficult problem and is beyond one’s capacity, hence notime should be wasted in thinking about it! This kind of “instantlearning” is detrimental to the acquirement of true understandingand it kills curiosity, thence along with it the pleasure of study.

3. Some contestants have the commendable habit of writing down notonly the mathematics, but remark on their progress as well. Somewould write down that they could not go from that point on, orwhat they did so far seemed to lead nowhere, or that they decidedto try a new approach. I really appreciate the manifestation ofthis kind of “academic sincerity”. (It is ironic to note that someleaders or deputy leaders tried to argue that those contestants were

43


almost near to the solution and should therefore be credited witha higher score. They might be, but they did not, and the goodintention of the leaders or deputy leaders is like filling in betweenthe lines for the contestants.) On the contrary, some of my studentsbehave in the opposite way. They write down the given in thefirst line (amounting to copying the first part of the question), andwrite down the conclusion at the end (amounting to copying thefinal part of the question), then fill in between with disconnectedpieces of information which may be relevant or irrelevant, endingwith an unfounded assertion “hence we conclude . . . ”! I am deeplydisappointed at this kind of insincerity, passing off gibberish as ananswer. I would have felt less disappointed if the student did notknow the answer at all.

I should add one more point about the “good” of mathematics competi-tions. A young friend of mine and a member of the Hong Kong 2012 IMOteam, Andy Loo, by recounting his own experience since primary schooldays with mathematics competitions, highlights the essential “good” ofmathematics competitions as lying in arousing a passion in the young-ster and piquing his or her interest in the subject. The experience ofparticipating in a mathematics competition can exert strong influenceon the future career of a youngster, whether he or she chooses to becomea research mathematician or not. For those who finally do not benefitfrom this experience for one reason or another, perhaps it is just an in-dication that they lack a genuine and sustained passion for the subjectof mathematics itself.

2 The “bad” of mathematics competitions

Despite what has been said in the previous section I have one worryabout mathematics competitions which has to do with the way of study-ing mathematics and doing mathematics, even more so for those who aredoing well in mathematics competitions. Those who do well in math-ematics competitions tend to develop a liking for solving problems byvery clever but sometimes quite ad hoc means, but lack the patience todo things in a systematic but hard way or view things in a more globalmanner. They tend to look for problems that are already well-posedfor them and they are not accustomed to dealing with vague situations.

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Pursuing mathematical research is not just to obtain a prescribed an-swer but to explore a situation in order to understand it as much asone can. It is far more important to be able to raise a good questionthan to be able to solve a problem set by somebody else who knows theanswer already. One may even change the problem (by imposing moreconditions or relaxing the demand) in order to make progress. This is,unfortunately, not what a contestant is allowed to do in a mathematicscompetition!

Of course, many strong contestants in various mathematics competitionsgo on to become outstanding mathematicians, but many stay at the levelof being good competition problem solvers even if they go on to pursuemathematics. Many leave the field altogether. That is not a problem initself, because everybody has his or her own aspiration and interest, andthere is no need for everybody to become a research mathematician. Onthe other hand, it would be a pity if they leave the field because they gettired of the subject or acquire a lopsided view of the subject as a resultof over-training during the youthful years they spent on mathematicscompetitions.

Looking at the history of several famed mathematics competitions we seea host of winners in the Eotvos Mathematics Competition of Hungary,started in 18942, went on to become eminent mathematicians [5]; we seemany medalists in the IMO’s, since the event started in 1959, received intheir subsequent career various awards for their important contributionto the field of mathematics, including the Fields Medal, NavanlinnaPrize, Wolf Prize, . . . [6]; we see the same happens for many PutnamFellows in the William Lowell Putnam Mathematical Competition in theUSA for undergraduates [8]. On the other hand, the Fields Medalist,Crafoord Prize and Wolf Prize recipient, YAU Shing-Tung, is noted forhis public view against mathematics competitions. The eminent Russianmathematician of the last century, Pavel Sergeevich Aleksandrov (1896–1982), was reported to have once said that he would not have becomethe mathematician he was had he joined the Mathematical Olympiad!An explanation of this polarity in opinions is to be sought in the way

2The role played by the journal Kozepiskolai Matematikai es Fizikai Lapok onmathematics and physics for secondary school founded in that same year meritsspecial attention. For more detailed information readers are invited to visit thewebsite of the journal [7].

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how one regards this activity known as a mathematics competition fromthe impression one gets in witnessing how it is run.

When I worked as a coordinator in the 35th IMO in 1994 I noticed thatsome teams scored rather high marks, but all six contestants in the teamworked out the problem in the same way, indicating solid training on theteam’s part. However, there were some teams, not all of whose membersscored as high marks, but each of whom approached the same problemin a different way, indicating a free and active mind that works indepen-dently and imaginatively. It made me wonder: will such qualities likeindependence and imagination be hampered by over-training, and if so,does that mean over-training for mathematics competitions defeats thepurpose of this otherwise meaningful activity? Rather than over-trainingwould an extended follow-up investigation of a competition problem en-able the youngsters to better appreciate what mathematical explorationis about? I am sure many contestants who go on to become outstandingmathematicians followed this practice of follow-up investigation duringthe youthful years they spent on mathematics competitions.

I will now illustrate with two examples. The first example is a ratherwell-known problem in one IMO. We will see how one can view it asmore than just a competition problem begging for just an answer. Theother example is on a research topic where the main problem is stillopen to this date (as far as I am aware of). We will see how a researchproblem differs from a problem viewed in the context of a mathematicscompetition problem.

It was natural that I paid some special attention to the questions set inthe 29th IMO, although I did not take part in the actual event in Julyof 1988. Question 6 of the 29th International Mathematical Olympiadreads:

Let a and b be positive integers such that ab + 1 divides a2 + b2.

Show thata2 + b2

ab+ 1is the square of an integer.

A slick solution to this problem, offered by a Bulgarian youngster(Emanouil Atanassov) who received a special prize for it, starts by sup-

posing that k = a2+b2

ab+1 is not a perfect square and rewriting the expres-

46


sion in the form

a2 − kab+ b2 = k, where k is a given positive integer. (∗)

Note that for any integral pair (a, b) satisfying (∗) we have ab ≥ 0, or elseab ≤ −1, and a2 + b2 = k(ab + 1) ≤ 0, implying that a = b = 0 so thatk = 0 ! Furthermore, since k is not a perfect square, we have ab > 0, thatis, none of a or b is 0. Let (a, b) be an integral pair satisfying (∗) witha > 0 (and hence b > 0) and a+ b smallest. We may assume a > b. (Bysymmetry we may assume a ≥ b. Note that a �= b or else k is a numberlying strictly between 1 and 2.) Regarding (∗) as a quadratic equationwith a positive root a and another root a′, we see that a+ a′ = kb andaa′ = b2 − k. Hence a′ is also an integer and (a′, b) is an integral pairsatisfying (*). Since a′b > 0 and b > 0, we have a′ > 0. But

a′ =b2 − k

a≤ b2 − 1

a≤ a2 − 1

a< a,

so that a′ + b < a + b, contradicting the choice of (a, b)! This proves

that a2+b2

ab+1 must be the square of an integer. (Having no access to theoriginal answer script I have tried to reconstruct the proof based on theinformation provided by a secondary source [1, p. 505]. The underlyingkey ideas are (i) choice of a minimal solution (a, b), and (ii) the expression(∗) viewed in the context of a quadratic equation.)

Slick as the proof is, it also invites a couple of queries.

1. What makes one suspect that a2+b2

ab+1 is the square of an integer?

2. The argument by reductio ad absurdum should hinge crucially uponthe condition that k is not a perfect square. In the proof thiscondition seems to have slipped in casually so that one does not seewhat really goes wrong if k is not a perfect square. More pertinently,

this proof by contradiction has not explained why a2+b2

ab+1 must be aperfect square, even though it confirms that it is so.

In contrast let us look at a less elegant solution, which is my own attempt.When I first heard of the problem, I was on a trip in Europe and had a“false insight” by putting a = N3 and b = N so that

a2 + b2 = N2(N4 + 1) = N2(ab+ 1).

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Under the impression that any integral solution (a, b, k) of k = a2+b2

ab+1 is

of the form (N3, N,N2) I formulated a strategy of trying to deduce froma2 + b2 = k(ab+ 1) the equality

(a−(3b2−3b+1)

)2+(b−1)2 =

(k−(2b−1)

)((a−(3b2−3b+1)

)(b−1)+1

).

Were I able to achieve that, then I could have reduced b in steps of oneuntil I got down to the equation a2+ b2 = k(ab+1) for which a = k = 1.By reversing steps I would have solved the problem. I tried to carry outthis strategy while I was travelling on a train, but to no avail. Uponreturning home I could resort to systematic brute-force checking andlook for some actual solutions, resulting in a (partial) list shown below.

a 1 8 27 30 64 112 125 216 240 343 418 512 . . .b 1 2 3 8 4 30 5 6 27 7 112 8 . . .k 1 4 9 4 16 4 25 36 9 49 4 64 . . .

Then I saw that my ill-fated strategy was doomed to failure, becausethere are solutions other than those of the form (N3, N,N2). However,not all was lost. When I stared at the pattern, I noticed that for a fixedk, the solutions could be obtained recursively as (ai, bi, ki) with

ai+1 = aiki − bi, bi+1 = ai, ki+1 = ki = k.

It remained to carry out the verification. Once that was done, allbecame clear. There is a set of “basic solutions” of the form (N3, N,N2)where N ∈ {1, 2, 3, . . . }. All other solutions are obtained from a “basicsolution” recursively as described above. In particular,

k =a2 + b2

ab+ 1

is the square of an integer. I feel that I understand the phenomenonmuch more than if I just learn from reading the slick proof.

[Thanks to Peter Shiu we can turn the indirect proof into a more trans-parent direct proof based on the same key ideas. Proceed as before andset c = min(a, b) and d = max(a, b). Look at the quadratic equation

x2 − kxc+ (c2 − k) = 0,

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for which d is a positive root with another root d′. Since d+d′ = kc anddd′ = c2 − k < c2 ≤ dc, we know that d′ is an integer less than c so thatd′ + c < 2c ≤ a + b. By the choice of (a, b), d′ cannot be positive. Onthe other hand,

(d+ 1)(d′ + 1) = dd′ + (d+ d′) + 1 = (c2 − k) + kc+ 1

= c2 + (c− 1)k + 1 ≥ c2 + 1 > 0.

Therefore, d′ + 1 > 0, implying that d′ = 0. Hence, k = c2 − dd′ = c2 isthe square of an integer.]

The next example is a research problem on the so-called Barker se-quence, which is a binary sequence of length S for which the sequenceand each off-phase shift of itself differ by at most one place of coincid-ing entries and non-coinciding entries in their overlapping part. Tech-nically speaking we say that the sequence has its aperiodic autocor-relation function having absolute value 0 or 1 at all off-phase values.For instance, 11101 is a Barker sequence of length 5, while 111010 oflength 6 is not a Barker sequence. Neither is 11101011 of length 8 aBarker sequence, but 11100010010 is a Barker sequence of length 11.For application in group synchronization digital systems in communi-cation science R. H. Barker first introduced the notion in 1953. Suchsequences for S equal to 2, 3, 4, 5, 7, 11, 13 were soon discovered and in1961 R. Turyn and J. Storer proved that there is no Barker sequenceof odd length S larger than 13. A well-known conjecture in combina-torial designs says that there is no Barker sequence of length S largerthan 13, which has withstood the effort of many able mathematiciansfor more than half a century. Although the conjecture itself remainsopen, it has stimulated much research in combinatorial designs and inthe design of sequences and arrays in communication science. In order tobetter understand the original problem researchers change the problemand look at the 2-dimensional analogue of arrays or even analogues inhigher dimensions, or other variations such as non-binary sequences andarrays over an alphabet set with more than two elements, or instead of asingle sequence a pair of sequences (known as Golay complementary se-quence pairs) satisfying some suitably formulated modification on theiraperiodic autocorrelation functions. In this sense the problem, insteadof looking like an interesting piece of curio, opens up new fields andgenerates new methods and techniques which prove useful elsewhere. (I

49


would recommend an excellent survey paper to those who wish to knowmore about this topic [2]).

3 School mathematics and “Olympiad mathemat-ics”

Since many mathematics competitions aim at testing the contestants’ability in problem solving rather than their acquaintance with specificsubject content knowledge, the problems are set in some general areaswhich can be made comprehensible to youngsters of that age group, in-dependent of different school syllabi in different countries and regions.That would cover topics in elementary number theory, algebra, combi-natorics, sequences, inequalities, functional equations, plane and solidgeometry and the like. Gradually the term “Olympiad mathematics”is coined to refer to this conglomeration of topics. One question that Iusually ponder over is this: why can’t this type of so-called “Olympiadmathematics” be made good use of in the classroom of school mathe-matics as well? If one aim of mathematics education is to let studentsknow what the subject is about and to arouse their interest in it, theninteresting non-routine problems should be able to play their part wellwhen used to supplement the day-to-day teaching and learning. In thepreface to Alice in Numberland: A Students’ Guide to the Enjoyment ofHigher Mathematics (1988) the authors, John Baylis and Rod Haggartyremark, “The professional mathematician will be familiar with the ideathat entertainment and serious intent are not incompatible: the problemfor us is to ensure that our readers will enjoy the entertainment but notmiss the mathematical point, . . . ”

By making use of “Olympiad mathematics” in the classroom I do notmean transplanting the competition problems directly there. Rather, Imean making use of the kind of topics, the spirit and the way the ques-tion is designed and formulated, even if the confine is to be within theofficial syllabus. The so-called “higher-order thinking” is (and shouldbe) one of the objectives in school mathematics as well. Sometimes wemay have underestimated the capability and the interest of our studentsin the classroom. It is not true that they only like routine (and henceusually regarded as “easy”?) material. Perhaps they lack the motivationto learn because they find the diluted content dull and are tired of it.

50


Besides, good questions not just benefit the learners in the classroom;it is also a challenging task for the teachers to design good questions,thereby upgrading themselves in the process. In this respect mathemat-ics competitions can benefit teachers as well, if they try to make use ofthe competition problems to enrich the learning experience of their stu-dents. To be able to better reap such benefit carefully designed seminarsfor teachers and suitably prepared didactical material will be helpful [3,pp. 1596–1597].

There is a well-known anecdote about the famous mathematician Johnvon Neumann (1903–1957). A friend of von Neumann once gave him aproblem to solve. Two cyclists A and B at a distance 20 miles apartwere approaching each other, each going at a speed of 10 miles per hour.A bee flew back and forth between A and B at a speed of 15 miles perhour, starting with A and back to A after meeting B, then back to Bafter meeting A, and so on. By the time the two cyclists met, how farhad the bee travelled? In a flash von Neumann gave the answer—15miles. His friend responded by saying that von Neumann must havealready known the trick so that he gave the answer so fast. His friendhad in mind the slick solution to this quickie, namely, that the cyclistsmet after one hour so that within that one hour the bee had travelled15 miles. To his friend’s astonishment von Neumann said that he knewno trick but simply summed an infinite series! (I leave it as an exercisefor you to find the answer by summing an infinite series.)

For me this anecdote is very instructive. For one thing, it tells me thatdifferent people may have different ways to go about solving a mathe-matical problem. There is no point in forcing everybody to solve it injust the same way you solve it. Likewise, there is no point in forcingeverybody to learn mathematics in just the same way you learn it. Thispoint dawned on me quite late in my teaching career. For a long timeI thought a geometric explanation would make my class understand lin-ear algebra in the easiest way, so I emphasized the geometric viewpointalong with an analytic explanation. I still continue to do that in class tothis date, but it did occur to me one day that some students may pre-fer an analytic explanation because they have difficulty with geometricvisualization. To von Neumann, who could carry out mental calculationwith lightning speed, maybe an infinite series was the first thing that

51


came up in his mind rather than the time spent by the cyclists meetingeach other!

Secondly, both methods of solution have their separate merits. Themethod of first calculating when the cyclists met is slick and capturesthe key point of the problem, killing it in one quick and direct shot.The other method of summing an infinite series, which is slower (butnot for von Neumann!) and is seemingly more cumbersome and not asclever, goes about solving the problem in a systematic manner, resortingeven to brute force. It indicates patience, determination, down-to-earthapproach and meticulous care. Besides, it can help to consolidate somebasic skills and nurture in a student a good working habit.

It makes me think that there are two approaches in doing mathematics.To give a military analogue one is like positional warfare and the otherguerrilla warfare. The first approach, which has been going on in theclassrooms of most schools and universities, is to present the subject ina systematically organized and carefully designed format supplementedwith exercises and problems. The other approach, which goes on morepredominantly in the training for mathematics competitions, is to con-front students with various kinds of problems and train them to look forpoints of attack, thereby accumulating a host of tricks and strategies.Each approach has its separate merit and they supplement and comple-ment each other. Just as in positional warfare flexibility and spontaneityare called for, while in guerrilla warfare careful prior preparation andgroundwork are needed, in the teaching and learning of mathematics weshould not just teach tricks and strategies to solve special types of prob-lems or just spend time on explaining the general theory and workingon problems that are amenable to routine means. We should let the twoapproaches supplement and complement each other in our classrooms.In the biography of the famous Chinese general and national hero of theSouthern Song Dynasty, Yue Fei (1103–1142) we find the description:“Setting up the battle formation is the routine of art of war. Manoeu-vring the battle formation skillfully rest solely with the mind.”

Sometimes the first approach may look quite plain and dull, comparedwith the excitement acquired from solving competition problems by thesecond approach. However, we should not overlook the significance ofthis seemingly bland approach, which can cover more general situations

52


and turns out to be much more powerful than an ad hoc method which,slick as it is, solves only a special case. Of course, it is true that frequentlya clever ad hoc method can develop into a powerful general method orcan become a part of a larger picture. A classic case in point is thedevelopment of calculus in history. In ancient time, only masters inmathematics could calculate the area and volume of certain geometricfigures, to name just a couple of them, Archimedes (c. 287 B.C.E. – c.212 B.C. E.) and Liu Hui (3rd century). Today we admire their ingenuitywhen we look at their clever solutions, but at the same time feel that itis rather beyond the capacity of an average student to do so. With thedevelopment of calculus since the seventeenth and eighteenth century,today even an average school pupil who has learnt the subject will haveno problem in calculating the area of many geometric figures.

Let me further illustrate with one example, which is a competitionproblem posed to me by the father of a contestant. In isosceles �ABC,where AB = AC and the measure of ∠BAC is 20◦, D is taken on theside AC such that AD = BC. Find θ, the measure of ∠ADB (see Figure1). Clearly, if one is to employ the law of sines, then the answer can be

Figure 1

readily obtained in a routine manner, namely,

AD

sin(α+ θ)=

AB

sin θ, AD = BC = 2AB sin

α

2,

53


where α is the measure of ∠BAC, thereby arriving at

tanα =sinα

2 sin α2 − cosα

.

When α = 20◦, θ = 150◦. However, the problem appeared in a pri-mary school mathematics competition in which the contestant was notexpected to possess the knowledge of the law of sines! Is there a way toavoid the use of this heavy machinery (for a primary school pupil)? I hitupon a solution by constructing an equilateral �FBC with F inside thegiven �ABC. Pick a point E on AB such that AE = CD (see Figure2).

Figure 2

Then it is not hard (by constructing DE, DF ) to find out that themeasure of ∠DBE is 10◦ (Exercise) so that θ = 180◦−20◦−10◦ = 150◦.Why would I throw in the equilateral�FBC as if by magic? It is becauseI had come across a similar-looking problem before: In an isosceles�ABC, where AB = AC and the measure of ∠BAC being 20◦, pointsD and E are taken on AC, AB respectively such that the measure of∠DBC is 70◦ and that of ∠ECB is 150◦; find φ, the measure of ∠BDE(see Figure 3). By constructing an equilateral �FBC with F inside the

54


Figure 3

given �ABC we can arrive at the answer φ = 10◦ (Exercise). These twoversions are indeed the description of the same situation, because it canbe proved that AD = BC in the second problem. Only knowledge ofcongruence triangles suffices. No knowledge of trigonometry is required.However, if the measure of ∠DBC and that of ∠ECB are not 70◦ and50◦ respectively, then the geometric proof completely breaks down! Butwe can still compute the measure of ∠BDE by employing the law ofsines, which is within what an average pupil learns in school. It hasto be admitted that the method is routine and not as elegant, but itcovers the general case and can be handled by an average pupil who hasacquired that piece of knowledge.

4 The pleasure (not pressure!) of mathematics com-petitions

Before working as a coordinator for the 35th IMO I harboured a distrustof the value of mathematics competitions. I still harbour this distrustto some extent, all the more when I witnessed during coordination ofthe IMO in 1994 how some leaders or deputy leaders over-reacted outof too much concern for winning high scores. Putting strong emphasison winning/losing will inculcate in the youngsters an unhealthy attitude

55


towards the whole activity. Attaching undue importance to the compe-tition by organizers, teachers, parents, students, is one main source thatmay cause distortion of the good intention of mathematics competitions,not to mention the more “commercial” consequences that take advan-tage of this misplaced emphasis. Not only does it fail to bring about theideal outcome of fostering genuine intrinsic interest and enthusiasm inthe subject, it takes the fun and meaning out of a truly extracurricularactivity as well. Instead of pleasure we are imposing pressure on theyoungsters.

Furthermore, the unilateral strengthening of ability to attain high scoreon these so-called “Olympiad mathematics” problems may have adverseeffect on the overall growth of a youngster, not just in terms of academicpursuit in other disciplines (or in mathematics itself!) but even in termsof personal development. In particular, I am disappointed at not findinghow mathematics competitions breathe life into a general mathematicsculture in the local scene. On the contrary, many people may be misledinto believing that those difficult “Olympiad mathematics” problemspresent the high point in mathematics, and that mathematics is thereforetoo difficult to lie within reach of an average person.

5 Concluding remark

On the whole I have great admiration for the talent of those youngsterswho take part in a mathematics competition. What little I accomplishin trying out those competition problems with all my might they accom-plish at a stroke, and explain it in a clear and lucid manner. I also havegreat respect for the dedication and enthusiasm of those organizers whobelieve in the value of a healthy mathematics competition. They areserving the mathematical community in many ways.

One predominant objection to mathematics competitions is the require-ment to work out the problems within a fixed time span, say three tofour hours. Some regard this as an act to undermine the intellectual andintrinsic pleasure of doing mathematics. In a comprehensive paper onmathematics competitions and mathematics education Petar Kenderovpoints out how this requirement disadvantages those creative youngsterswho are “slow workers”. Along with it he points out some important

56


features which are not encouraged in a traditional mathematics com-petition but which are essential for doing good work in mathematics.These include “the ability to formulate questions and pose problems,to generate, evaluate, and reject conjectures, to come up with new andnon-standard ideas”. Moreover, he points out that all such activities“require ample thinking time, access to information sources in librariesor the internet, communication with peers and experts working on simi-lar problems, none of which are allowed in traditional competitions.” [3,p. 1592]

My good friend, Tony Gardiner, who is known for his rich experience inmathematics competitions and had served as the leader of the BritishIMO team four times, after reading my article in 1995 [4] commentedthat I should not blame the negative aspects on the mathematics compe-tition itself. He went on to enlighten me on one point, namely, a math-ematics competition should be seen as just the tip of a very large, moreinteresting, iceberg, for it should provide an incentive for each countryto establish a pyramid of activities for masses of interested students. Itwould be to the benefit of all to think about what other activities be-sides mathematics competitions can be organized to go along with it.These may include the setting up of a mathematics club or publishing amagazine to let interested youngsters share their enthusiasm and theirideas, organizing a problem session, holding contests in doing projectsat various levels and to various depth, writing book reports and essays,producing cartoons, videos, softwares, toys, games, puzzles, . . . . I wishmore people will see mathematics competitions in this light, in whichcase the negative impression, which I might have conveyed in this paper,will no longer linger on!

The good, the bad and the pleasure of mathematics competitionsAre to which we should pay our attention.

Benefit from the good; avoid the bad;And soak in the pleasure.

Then we will find for ourselves satisfaction!

References

[1] Djukic, D., Matic, I., Jankovic, V., Petrovic, N., The IMO Com-pendium: A Collection of Problems Suggested for the International

57


Mathematical Olympiads, 1959–2004, New York: Springer-Verlag,2006.

[2] Jedwab, J., What can be used instead of a Barker sequence? Con-temporary Mathematics, 461 (2008), pp. 153–178.)

[3] Kenderov, P.S., Competition and mathematics education, in Pro-ceedings of the International Congress of Mathematicians: Madrid,August 22–30, 2006, edited by Marta Sanz-Sole et al, Zurich: Euro-pean Mathematical Society, 2007, pp. 1583–1598.

[4] Siu, M. K., Some reflections of a coordinator on the IMO, Mathe-matics Competitions, 8(1) (1995), pp. 73–77.

[5] http://www.batmath.it/matematica/raccolte_es/ek_competitions/ek_competitions.pdf

[6] http://www.imo-official.org/

[7] http://www.komal.hu/info/bemutatkozas.h.shtml

[8] http://www.maa.org/awards/putnam.html

Siu Man Keung

Department of Mathematics

University of Hong Kong

Hong Kong SAR

CHINA


58


The Twin Towers of Hanoi

Jack Chen & Richard Mah & Steven Xia

In a famous puzzle known as the Tower of Hanoi, there are three pegsin the playing board. There are n disks of different sizes, all stacked onthe first peg, in ascending order of size from the top. The objective is totransfer this tower to the third peg. The rule is that we may only moveone disk on top of a peg to the top of another peg, and a disk may notbe placed on top of a smaller disk. Figure 1 shows the starting positionwhen n = 3.

CBA

........................................

........................................

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.......

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.......

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Figure 1

If n = 1, the task can be accomplished in 1 move: A3. If n = 2, thetask can be accomplished in 3 moves: A2BA3. If n = 3, the task can beaccomplished in 7 moves:: A3BA2CA1BA3. The subscript for A showsthe number of the peg to which it is moving. B and C never have choices.

Based on these three simple cases, we conjecture that the minimumnumber of moves required to transfer a tower with n disks is 2n− 1. Letus prove this by mathematical induction. We assume that the minimumnumber of moves required to transfer a tower with n disks is 2n − 1. Wenow try to transfer a tower with n+ 1 disks.

A critical moment occurs when the bottom disk is moving, from the firstpeg to the third. In order for this to be possible, the n smaller disks mustform a tower on the second peg. Thus before the move of the bottomdisk, we must transfer a tower with n disks from the first peg to thesecond. By the induction hypothesis, this takes 2n − 1 moves. After themove of the bottom disk, we must complete the task by transferring thetower on the second peg, consisting of the n smaller disks, to the thirdpeg. Once again, this also takes 2n−1 moves. Together with the move of

59


the bottom disk, the minimum number is (2n−1)+1+(2n−1) = 2n+1−1.This completes the inductive argument.

We can express the solution to the problem of the Tower of Hanoi usinga different terminology. Let an be the minimum number of moves totransfer a tower of height n from one peg to another. From our analysisabove, we see that an = 2an−1+1. This result, which defines an in termsof an−1, is an example of what is called a recurrence relation. Alongwith a1 = 1, which is called an initial value, they define the sequence{an} uniquely.

Using the recurrence relation and the initial value, we can generateadditional terms of the sequence, as shown in the chart below.

a1 = 1 a2 = 3 a3 = 7

a4 = 15 a5 = 31 a6 = 63

a7 = 127 a8 = 255 a9 = 511

We now consider a variant which we call the Twin Towers of Hanoi. Asbefore, there are three pegs in the playing board. There are n disks ofsizes 1, 2, 3, . . . , n. Those of odd sizes are stacked on the first peg, andthose of even sizes are stacked on the third peg. On both pegs, the disksare in ascending order of size from the top. The rule is the same, in thatwe may only move a disk on top of a peg to the top of another peg, anda disk may not be placed on top of a smaller disk. Figure 2 illustratesthe starting position of the case n = 6.

531 2

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Figure 2

The objective is to have the two towers trade places. As in the Tower ofHanoi, a critical moment occurs when disk 6 is moving, from the thirdpeg to the first. In order for this move to be possible, the 5 smaller disks

60


must form a tower on the second peg, as illustrated in Figure 3. Thusbefore the move of disk 6, we must merge the disks 1 to 5 into a toweron the second peg.

12345 6........................................

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Figure 3

We have identified an intermediate objective for the Twin Towers ofHanoi, that of merging the n disks on the second peg. Figure 4 illustratesthe case n = 5.

135 4

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1

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Figure 4

We first merge disks 1 and 2 on the second peg to pave the way for themove of disk 3. After disk 3 moves on top of disk 4, we transfer disks1 and 2 on top of disk 3 to pave the way for the move of disk 5. Afterdisk 5 moves to the second peg, we transfer disks 1, 2, 3 and 4 on top ofit to complete the task.

In general, we merge disks 1, 2, 3, . . . , n − 3 on the second peg, movedisk n− 2 on top of disk n− 1, transfer disks 1, 2, 3, . . . , n− 3 on topof disk n − 2, move disk n to the second peg and transfer disks 1, 2, 3,. . . , n− 1 on top of disk n.

This means that if we let bn be the minimum number of moves formerging n disks, then we have bn = bn−3 + 1 + an−3 + 1 + an−1. Sincean = 2n − 1, we have bn = bn−3 + 5 · 2n−3.

This is a three-step recurrence relation since bn is not defined in termsof bn−1 but in terms of bn−3. Hence we need three initial values. It isnot hard to determine b1, b2 and b3, and generate the chart of valuesbelow.

b1 = 1 b2 = 2 b3 = 5

b4 = 11 b5 = 22 b6 = 45

b7 = 91 b8 = 182 b9 = 365

62


We notice that bn is either equal to 2bn−1 or 2bn−1 + 1, but it is notimmediately clear what the general formula for bn is.

We also notice that the numbers in the last column so far are all multiplesof 5. Dividing them by 5 yields the quotients 1, 9 and 73. In the table ofvalues of an given earlier, we notice that the numbers in the last columnthere are all multiples of 7. Dividing them by 7 yields the quotients 1, 9and 73 also. Thus we suspect that bn is roughly 5

7an.

57a1 = 1− 2

757a2 = 2 + 1

757a3 = 5

57a4 = 11− 2

757a5 = 22 + 1

757a6 = 45

57a7 = 91− 2

757a8 = 182 + 1

757a9 = 365

Thus we conjecture that bn is equal to 57an rounded off to the nearest

integer. More specifically, we conjecture that

bn =

57 (2

n − 1) when n ≡ 0 (mod 3),57 (2

n − 2) + 1 when n ≡ 1 (mod 3),57 (2

n − 4) + 2 when n ≡ 2 (mod 3).

We now use mathematical induction to prove these three formulae. First,let n ≡ 0 (mod 3). We have 5

7 (23− 1) = 5 = b3. Suppose bn = 5

7 (2n− 1)

for some n ≥ 3. Then

bn+3 = bn − 5 · 2n

= 57 (2

n − 1) + 5 · 2n

= 57 (2

n − 1 + 7 · 2n)= 5

7 (2n+3 − 1).

The other two formulae can be proved in an analogous manner.

Returning to the main task of having the two towers trade places. Let theminimum number of moves required be cn where n is the total numberof disks. We have already identified the critical moment when disk nmoves, either from the first peg to the third or vice versa. In order forthis move to be possible, the n − 1 smaller disks must form a tower onthe second peg. This merger requires bn−1 moves. After the move of

63


disk n, we must disperse the merged tower. This also takes bn−1 movessince the process is reversible. It follows that cn = 2bn−1 + 1. We havethe chart of values below.

c1 = 1 c2 = 3 c3 = 5

c4 = 11 c5 = 23 c6 = 45

c7 = 91 c8 = 183 c9 = 365

It follows that cn and bn are identical, except for n ≡ 2 (mod 3), whenthey differ by 1. It may be interesting to discover what may have causedthis discrepancy.

The method of mathematical induction allows us to prove that thespecified minimum numbers of moves are indeed correct, but it doesnot explain why they should have those values. Moreover, we have tohave an idea what these values are before we can apply the inductiveargument. We have resorted to observing patterns and making inspiredguesses. We would like to have a more formal and systematic approachto solving recurrence relations.

Let us revisit an = 2an−1 +1 with initial value a1 = 1. Actually, a0 = 0will serve just as well. Note that the recurrence relation is true notjust for one value of n, but for all values of n. In other words, all thestatements below are true. This is called an iteration.

an = 2an−1 + 1,

an−1 = 2an−2 + 1,

an−2 = 2an−3 + 1,

. . .

a2 = 2a1 + 1,

a1 = 2a0 + 1.

Multiplying each by a power of 2 one higher than the preceding one, we

64


obtain the following statements.

an = 2an−1 + 1,

2an−1 = 22an−2 + 2,

22an−2 = 23an−3 + 22,

. . .

2n−2a2 = 2n−1a1 + 2n−2,

2n−1a1 = 2na0 + 2n−1.

Addition results in massive cancellations, leading to

an = 2na0 + 1 + 2 + 22 + · · ·+ 2n−2 + 2n−1 = 2n − 1.

We now turn to the recurrence relation bn = bn−3 + 5 · 2n−3 with initialvalues b0 = 0, b1 = 1 and b2 = 2. We consider only n ≡ 0 (mod 3) asthe other two cases are analogous.

bn = bn−3 + 5 · 2n−3,

bn−3 = bn−6 + 5 · 2n−6,

bn−6 = bn−9 + 5 · 2n−9,

· · · = · · · ,b6 = b3 + 5 · 23,b3 = b0 + 5 · 20.

Addition yields

bn = b0 + 5(1 + 23 + 26 + · · ·+ 2n−3)

=5(2n − 1)

23 − 1=

5

7(2n − 1).

There is an alternative approach which is equally formal and systematic.Rewriting this recurrence relation as an − 2an−1 = 1, we consider theassociated recurrence relation an − 2an−1 = 0. The simpler one is saidto be homogeneous because all terms involve the sequence {an}. Theoriginal recurrence relation is said to be non-homogeneous since it has anon-zero term not involving the sequence {an}.

65


We may solve the homogeneous recurrence relation by setting an = xn

for some non-zero number x. Then xn−2xn−1 = 0. Since x �= 0, we cansimplify to x− 2 = 0 so that an = 2n is a solution to the homogeneousrecurrence relation. The general solution is an = K2n for some constantK. This value will eventually be determined by the initial value.

All we need now is a particular solution to the non-homogeneous recur-rence relation. Since the non-homogeneous term is 1, it is reasonable toguess that an = A for some constant A. It is called an (as yet) undeter-mined coefficient. Now 0 = an − 2an−1 = A− 2A = −A. Hence A = −1and the particular solution is an = −1.

Combining the two partial solutions, we now have the general solutionto the non-homogeneous recurrence relation, namely, an = K2n − 1.Setting n = 0, we have 0 = a0 = K20 − 1 = K − 1. It follows thatK = 1, so that an = 2n − 1.

We now turn to the recurrence relation bn−bn−3 = 5·2n. The associatedhomogeneous recurrence relation is bn − bn−3 = 0. Setting bn = xn, wehave xn − xn−3 = 0 so that

0 = x3 − 1 = (x− 1)(x2 + x+ 1).

The first factor yields the root x = 1 while the second factor yields com-

plex roots 1±√3i

2 , where i is a root of the quadratic equation x2 + 1 = 0.

We denote 1+√3i

2 by ω. It satisfies ω2 = 1−√3i

2 , ω3 = 1 and ω2+ω+1 = 0.Thus the general solution of the aassociated homogeneous recurrence re-lation is bn = K1 +K2ω

n +K3ω2n.

We need a particular solution to the non-homogeneous recurrence rela-tion. Since the non-homogeneous term is 5 · 2n, it is reasonable to guessthat bn = A2n for some constant A. Then A2n = A2n−3 + 5 · 2n−3.Canceling the factor 2n−3 yields 8A = A + 5. Hence A = 5

7 so thatbn = 5

72n.

It follows that the general solution to the non-homogeneous recurrencerelation is bn = 5

72n + K1 + K2ω

n + K3ω2n. Using the initial values

66


b0 = 0, b1 = 1 and b2 = 2, we have

0 = b0 =5

7+K1 +K2 +K3,

1 = b1 =10

7+K1 +K2ω +K3ω

2,

2 = b2 =20

7+K1 +K2ω

2 +K3ω.

It follows that

K1 + K2 + K3 = −5

7,

K1 + ωK2 + ω2K3 = −3

7,

K1 + ω2K2 + ωK3 = −6

7.

We can solve this system of three equations in three unknowns and obtainK1 = − 2

3 , K2 = − 221 − 1

7ω and K3 = − 221 − 1

7ω2. However, we do not

really need these values.

For n ≡ 0 (mod 3), we have

bn =5

72n +K1 +K2 +K3

=5

72n − 5

7

=5

7(2n − 1).


bn =5

72n +K1 +K2ω +K3ω

2

=5

72n − 3

7

=5

7(2n − 2) + 1.

67



bn =5

72n +K1 +K2ω

2 +K3ω

=5

72n − 6

7

=5

7(2n − 4) + 2.

Steven XiaGrade 9Vernon Barford Junior High School32 Fairway Drive NWEdmonton, Alberta, [email protected]

Jack ChenGrade 8Vernon Barford Junior High School32 Fairway Drive NWEdmonton, Alberta, [email protected]

Richard MahGrade 8Vernon Barford Junior High School32 Fairway Drive NWEdmonton, Alberta, [email protected]

68


Tournament of the TownsSelected Problems, Spring 2012

Andy Liu

1. Five students have the first names: Clark, Donald, Jack, Robinand Steven, and have the last names, in a different order: Clarkson,Donaldson, Jackson, Robinson and Stevenson. Clark is 1 year olderthan Clarkson, Donald is 2 years older than Donaldson, Jack is 3years older than Jackson and Robin is 4 years older than Robinson.Who is older, Steven or Stevenson and what is the difference intheir ages?

Solution:Let Steven be n years older than Stevenson. The total age of Clark,Donald, Jack, Robin and Steven must be the same as the totalage of Clarkson, Donaldson, Jackson, Robinson and Stevenson,because these are the same five people. Hence 1+2+3+4+n = 0so that n = −10. It means that Steven is 10 years younger thanStevenson.

2. The game Minesweeper is played on a 10 × 10 board. Each celleither contains a bomb or is vacant. On each vacant cell is recordedthe number of bombs in the neighbouring cells along a row, acolumn or a diagonal. Then all the bombs are removed, and newbombs are placed in all cells which were previously vacant, and thenumbers of neighbours are recorded as before. Can the sum of allnumbers on the board now be greater than the sum of all numberson the board before?

Solution:Represent each cell by a vertex and join two neighbouring cells byan edge. An edge is called a scoring edge if it joins a bomb cell andvacant cell, since it will contribute 1 to the number recorded on thevacant cell. An edge which joins two bomb cells or two vacant cellsis non-scoring. Hence the sum of the recorded numbers is equalto the number of scoring edges. After the transformation of theboard, a scoring edge remains a scoring edge, and a non-scoring

69



Andy Liu





69



Andy Liu





69


edge remains a non-scoring edge. Hence the sum of all the numberson the board before must be equal to the sum of all the numberson the board after.

3. A circle touches sides AB, BC and CD of a parallelogram ABCDat points K, L and M respectively. Prove that the line KL bisectsthe perpendicular dropped from C to AB.

Solution:Let KL intersect the perpendicular CH dropped from C to ABat P . Let the extensions of KL and DC intersect at Q. Now∠KLM = 90◦ since KM is a diameter of the circle. Hence QLMis a right triangle, so that its circumcentre lies on QM as well as onthe perpendicular bisector of LM . Now CL = CM since both aretangents from C to the circle. Hence C lies on the perpendicularbisector of LM . Being on QM , C must be the circumcentre oftriangle QLM . This means that QC = CM . Now PC is parallelto KM . By the Midpoint Theorem, PC = 1

2KM = 12HC, which

is equivalent to the desired result.

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B H K A

Q C M D

P

L

4. Among 239 coins which look the same, there are two counterfeitcoins of the same weight, and 237 real coins of the same weightbut different from that of the counterfeit coins. Determine in threeweighings on a balance whether the counterfeit coins are heavieror lighter than the real coins. It is not necessary to identify thecounterfeit coins.

Solution:Put 80 coins into group A, 79 coins into each of groups B and C,and we have one coin left. If we add this coin to group B, we call

70


the expanded group B+. C+ is similarly obtained from C. In thefirst two weighings, we try to balance group A against group B+

and group A against group C+. We consider three cases.Case 1. We have equilibrium both times.This means that each group has one counterfeit coin, so that theextra coin must be counterfeit. The coins in both groups B and Care real. Weigh one of them against the known counterfeit coin,and that will tell us the desired answer.Case 2. We have equilibrium only once, say between A and B+.Either each has a counterfeit or neither has a counterfeit coin.Divide the coins in A into two subgroups of 40 and weigh themagainst each other. If we have equilibrium, A does not have anycounterfeit coins, and neither does B, which means C has both ofthem. If we do not have equilibrium, A has a counterfeit coin, andso does B+. This means that the extra coin is real and C+ doesnot have any counterfeit coins. In either situation, the weighingbetween A and C+ tells us the desired answer.Case 3. We have no equilibrium.We claim that either A is heavy both times or A is light bothtimes. Suppose to the contrary that B+ is heavier than A andA is heavier than C+. Then they must have two, one and zerocounterfeit coins in either order. However, if A has one counterfeitcoin, it is impossible for either B or C to have two of them. Thisjustifies our claim. So either both B+ and C+ have no counterfeitcoins or both have one. Divide the coins in B+ into two subgroupsof 40 and weigh them against each other. This will tell us thedesired answer.

5. An infinite sequence of numbers a1, a2, a3, . . . is given. For anypositive integer k, there exists a positive integer t = t(k) such thatak = ak+t = ak+2t = . . . Is this sequence necessarily periodic?

Solution:The result is not true, and here is a counter-example. For anypositive integer k, let ak be the highest power of 2 which dividesk. Thus the sequence is 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, . . . Thechoice t(k) = 2ak+1 satisfies the given condition and yet the se-quence is not periodic as the maximum size of the terms increaseswithout bound.

71


6. I is the incentre of triangle ABC. A circle through B and Iintersects AB at F and BC at D. K is the midpoint of DF .Prove that ∠AKC is obtuse.

Solution:Let ACDF be a convex quadrilateral. Let K and M be therespective midpoints of DF and AC. We claim that KM <AF+CD

2 . Let L be the midpoint of AD, as shown in the diagram

below on the left. By the Midpoint Theorem, KL = AF2 and

LM = CD2 . By the Triangle Inequality, KM < KL + LM =

AF+CD2 , justifying our claim.

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A M C

L

F

KD

A E C

F

B

D

I

Let E be the point on AC such that AE = AF , as shown in thediagram above on the right. Let ∠AEI = α and ∠CEI = β,so that α + β = 180◦. Since AI bisects ∠CAB, triangles AEIand AFI are congruent. Hence ∠AFI = α and ∠BFI = β.Since BDIF is cyclic, ∠BDI = α and ∠CDI = β = ∠CEI.Hence triangles CDI and CEI are also congruent, so that wehave CD = CE. Let M be the midpoint of AC. By our earlierclaim, KM < AF+CD

2 = AE+CE2 = AC

2 . Hence K lies within thesemicircle with diameter AC, which implies that ∠AKC is obtuse.

7. Peter chooses some positive integer a and Paul wants to determineit. Paul only knows that the sum of the digits of Peter’s number is2012. In each moves, Paul chooses a positive integer x and Petertells him the sum of the digits of |x − a|. What is the minimalnumber of moves Paul needs to determine Peter’s number for sure?

Solution:We claim that Paul needs at most 2012 questions. Let Peter’snumber have n non-zero digits a1, a2, . . . , an from right to left,with n > 1 and a1 + a2 + · · ·+ an = 2012. For 1 ≤ k ≤ n, let bk be

72


the number of adjacent 0s to the right of ak. Paul will carry out thefollowing steps. In Step 1, Paul determines b1 using one question.He chooses 1, and Peter’s answer must be 2011 + 9b1. In step k,2 ≤ k ≤ n, Paul will determine ak−1 and bk using ak−1 questions.Paul chooses a number which has a 2 in front of the known digitsof Peter’s number. If Peter’s answer is 2010, Paul replaces the2 by the 3. Continuing this way, when the first digit of Paul’schoice is ak−1, Peter will answer 2013−ak−1. However, when Paulreplaces ak−1 by ak−1+1, Peter’s answer will be 2012−ak−1+9bk.Now only an is left, but Paul does not know that unless an = 1.However, by the time he asks the (an−1)-st questions here, he willget 0 as a response and knows Peter’s number. The total numberof questions needed is 1 + a1 + a2 + · · · + (an − 1) = 2012. Thisjustifies our claim. We now prove that Peter can force Paul to use2012 questions, by constructing his number one step at a time. Letb1 be the number of digits in Paul’s first choice. Peter chooses thelast b1 digits of his number to be 0 and put a 1 in front, referredto as a1. His response will be 2011 + 9b1. Paul can only tell thatPeter’s number ends in exactly b1 zeros. Paul’s next choice mustbe a number with more digits. Peter adds b2 zeros in front of a1to match the length of Paul’s choice, and put another 1 in front,referred to as a2. Paul can only determine the last b2+1+b1 digitsof Peter’s number, plus the fact that the digit in front is non-zero.Continuing this way, we have a1 = a2 = · · · = a2012 = 1, and Paulneeds one question to determine each of b1, b2, . . . , b2012.

Andy Liu

University of Alberta

CANADA


73

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